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\lhead{MAT\,137\,Y1Y}
\chead{Problem Set \#\,1}
\rhead{Summer 2014}

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\begin{large}
  \noindent
  Name : Robert Staskiewicz \hfill Student Number: 1000340570\\
\end{large}

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\subsection*{Topic: Sets, Domains, and Ranges}

\medskip

\begin{enumerate}
 
       
    % place solution to question 1 below

    \item
    \medskip
    \noindent
    Consider the following claim:
    \begin{quote}
    If A, B and C are sets, prove that $B \setminus (A \cup C) = (B \setminus A) \cap (B \setminus C).$
    \end{quote}
    We can express this in the notation of symbolic logic as:
    $$ \ \forall \ Sets \ A,B,C,\ [B \setminus (A \cup C)] = [(B \setminus A) \cap (B \setminus C)] $$
    {\bf Proof:} \\
    \begin{pindent}
		
	\hspace*{-10mm}	We must Prove: $[(B\setminus(A \cup C)) \subseteq ((B \setminus A) \cap (B \setminus C))] \land [((B \setminus A) \cap (B \setminus C)) \subseteq (B\setminus(A \cup C))]$\\
		
        \begin{assumption}{$x \in (B \setminus (A \cup C))$}
		Then $x\in B \land x \notin (A \cup C)$\\
		Then $x \in B \land x \notin A \land x \notin C$ \just{x is not a member of $(A \cup C)$.}\\
		Then $x\in (B \setminus A)$ \just{$x \in B \land x \notin A$}\\
		Then $x\in (B \setminus C)$ \just{$x \in B \land x \notin C$}\\
		Then $x \in  ((B \setminus A) \cap (B \setminus C)) $ \just{$x\in (B \setminus A) \land x\in (B \setminus C)$ }
		\end{assumption}
		Therefore, $(B\setminus(A \cup C)) \subseteq ((B \setminus A) \cap (B \setminus C)) $\\
		
		\begin{assumption}{$x \in  ((B \setminus A) \cap (B \setminus C))$}
				Then $(x \in B \land x \notin A) \land (x \in B \land x \notin C) $ \just{Definition of set intersection.} \\
				Case1:\\
				$x \in B \land x \notin A  $ \\
				Then $x \in (B \setminus (A \cup C)) $\just {x is in B and not in A}\\
				Case2:\\
				$x \in B \land x \notin C$\\
				Then $x \in (B \setminus (A \cup C)) $\just {x is in B and not in C}
		\end{assumption}
		Therefore $ ((B \setminus A) \cup (B \setminus C)) \subseteq (B\setminus(A \cup C))$\\\\
	\hspace*{-10mm}Therefore, $ \ \forall \ Sets \ A,B,C,\ [B \setminus (A \cup C)] = [(B \setminus A) \cap (B \setminus C)] $
    \end{pindent}
    \medskip
    
    % place solution to question 2 below

       
    \item
    \begin{enumerate}
    \item
    Consider the following claim:
       \begin{quote} 
       "A function $f$ is periodic if and only if there exists a $k > 0$ such that for every $x,f(k + x) = f(x)$".\\
       \end{quote}
       We can express this in the notation of symbolic logic as:
       $$    \text{A function is periodic} \iff \exists k \in \R, k>0, \forall x \in \R, f(k+x) = f(x)  $$ \\
       
       \item
       
       \end{enumerate}
       
           Consider the following claim:
           \begin{quote}
           Prove the following by contradiction: If $||x|-|y||<|x-y|, \text{then}\  xy < 0$.
           \end{quote}
       {\bf Proof:} \\
    We can express this in the notation of symbolic logic as:
    $$\forall a \in \Z, [(n-1) \land n \land (n+1)] \implies [(n+1)^3 \not= [n^3 + (n-1)^3]] $$ \\
		{\bf Proof:} \\
		    \begin{pindent}
		
          \begin{assumption}{$(n+1)^3=n^3+(n-1)^3$}
				Then $(n+1)^3=n^3+3n^2+3n+1$\\
				And $n^3+(n-1)^3=n^3+n^3-3n^2+3n-1$\\
				Then $n^3+3n^2+3n+1= n^3+n^3-3n^2+3n-1$\\
				Then $n^3+3n^2+3n+1= 2n^3-3n^2+3n-1$\\
				Then $n^3-6n^2-2=0$\\
				Then $n^2(n-6)-2=0$\\
				Then $n^2(n-6)=2$\\
				Then $(n^2>0) \land ((n-6)>0)$ \just{Both products mut be positive.}\\
				Then $n^2>0$\ is always true. \just{The square of any number is positive.}\\
				Then we have $(n-6)>0$ \just{This term must be positive for the product to be positive.}
				Then $n>6$ \just{Adding 6 to both sides. }\\
				Then $(n>6) \implies [n^2>36)]$\\
				But $[(n^2>36)\land (n>6)] \implies [n^2(n-6) \not=2]$\\
				We reach a contradiction. Thus, our original implication is true.
          \end{assumption}
          Therefore,$\forall a \in \Z, [(n-1) \land n \land (n+1)] \implies [(n+1)^3 \not= [n^3 + (n-1)^3]] $
           
      \end{pindent}
      \medskip

	\item
	\item
	\item
	\item
		\begin{enumerate}
		\item
		\item
		 Consider the following function:\\
		 \begin{quote}
		  Domain $=  (-\infty, 1)\cup (1, 3) \cup (3, \infty) $ and a Range $= (5, \infty) $\\\\
		  A function with this Domain and Range is: $ f(x) = (\frac{1}{x^2}+5)\frac{x^2-4x+3}{(x-1)(x-3)} $\\
		  
		  \end{quote}
		  	{\bf Proof:} \\
		  		    \begin{pindent}
		  The Domain appears to be defined at all points but $1$ and $3$.\\
		  We take a function that has an undefined denominator at $x=1$ and $x=3$,
		  so we have $(x-3)$ and $(x-1)$ for a denominator $\frac{1}{(x-1)(x-3)}$. However, we want these points to be undefined, so we make the numerator $(x-3)(x-1)$.
		\end{pindent}
		      \medskip
		\end{enumerate}
\end{enumerate}

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